Problem: $f\,^{\prime}(x)=-\dfrac{320}{x^6}$ and $f(1)=30$. $f(-2) = $
Finding $f(x)$ We have $f'(x)=-\dfrac{320}{x^6}$ and we want to find $f(x)$ : $\begin{aligned}f(x)& = \int f'(x)\,dx \\\\ & = \int (-\dfrac{320}{x^6})\,dx \\\\ & = {64x^{-5}} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(1)=30$. Here's what we get when we plug in $1$ : $\begin{aligned}f(1)&={64(1)^{-5}} {+ C}\\\\ &={64} {+ C} \end{aligned}$ We are given that this must equal $30$ : $30 = {64} {+ C}$ Solving the equation gives us ${C=-34}$. Finding $f(-2)$ Now, we have that $f(x)= {64x^{-5}} {-34}$. Let's find $f(-2)$ by plugging in $-2$ : $\begin{aligned}f(-2)&=64(-2)^{-5}-34\\\\ &=-36 \end{aligned}$ The answer $f(-2) = -36$